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Complex Analysis : Residue Theory

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = \frac{e^{-z}}{z^3}$

in the region |z|=4 .

Possible Answers:

$- 2 \pi i$

$- \frac{\pi i }{4}$

$3 \pi i$

$- \frac{\pi i }{2}$

$\pi i$

Correct answer:

$\pi i$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{z^3}$

a singularity exists where z= 0 . Thus, since where z=0 is the only singularity for f(z) inside |z|=3 , we seek to evaluate the residue for z=0 .

Observe,

$f(z) \newline = \frac{e^{-z}}{z^3} \newline = \frac{1}{z^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^{k-3}}{k!}$

The coefficient of $\frac{1}{z}$ is $\frac{(-1)^{2}}{2!} = \frac{1}{2}$ .

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{z^3}] = \frac{1}{2}$ .

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{z^2}dz = 2 \pi i \text{Res}_{z=0}[\frac{e^{-z}}{z^2}] = 2 \pi i (\frac{1}{2}) = \pi i$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{z^2} dz = \pi i$

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Using Cauchy's Residue Theorem, evaluate the integral of

$f(z) = \frac{e^{-z}}{(z-1)^3}$

in the region |z|=4

Possible Answers:

$\frac{ \pi i}{e}$

$- 2 \pi i$

$-e^{2 \pi i}$

$2 \pi i$

$- 2 \pi i(1 + e)$

Correct answer:

$\frac{ \pi i}{e}$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{(z-1)^3}$

a singularity exists where z= 1 . Thus, since where z=1 is the only singularity for f(z) inside |z|=4 , we seek to evaluate the residue for z=1 .

Observe,

$f(z) \newline = \frac{e^{-z}}{(z-1)^3} \newline = \frac{e^{-u-1}}{u^3} \newline = \frac{ 1}{e} * \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^k}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^{k-3}}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{-1^{k}(z-1)^{k-3}}{k!}$

The coefficient of $\frac{1}{z-1}$ is $\frac{1}{e} * \frac{(-1)^{2}}{2!} = \frac{1}{2e}$ .

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{(z-1)^2}] = \frac{1}{2e}$ .

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{(z-1)^2}dz = 2 \pi i \text{Res}_{z=1}[\frac{e^{-z}}{(z-1)^2}] = 2 \pi i (\frac{1}{2e}) = \frac{ \pi i}{e}$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{(z-1)^2} dz = \frac{ \pi i}{e}$

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = z^3 e^{\frac{1}{z}}$

in the region |z|=4 .

Possible Answers:

$2 \pi i$

$- \frac{\pi i }{6}$

$\frac{\pi i}{4}$

$\frac{\pi i}{12}$

$\frac{\pi i }{24}$

Correct answer:

$\frac{\pi i}{12}$

Explanation:

Note, there is one singularity for f(z) where z = 0 .

Let

$u = \frac{1}{z}$

Then

$z = \frac{1}{u}$

$f(z) = z^3 e^{\frac{1}{z}} = \frac{1}{u^3} e^{u}$ .

Therefore, there is one singularity for f(z) where u = 0 . Hence, we seek to compute the residue for f(z) where u = 0 = z

Observe,

$f(z) \newline = z^{3} e^{\frac{1}{z}} \newline = \frac{1}{u^3} e^u \newline = \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{u^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{u^{k-3}}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{z^{3-k}}{k!}$

So, when k = 4 , $z^{3-k} = z^{-1} = \frac{1}{z}$ .

Thus, the coefficient of $\frac{1}{z}$ is $\frac{1}{4!}$ .

Therefore,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0} [z^2 e^{\frac{1}{z}}] = \frac{1}{4!}$

Hence, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = 2 \pi i \text{Res}_{z=0}[z^2 e^{\frac{1}{z}}] = 2 \pi i * \frac{1}{4!} = \frac{\pi i}{12}$

Therefore,

$\displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = \frac{\pi i}{12}$

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Example Question #2 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})]$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use the Residue Theorem to evaluate the integral of

$f(z) = \frac{z^6}{1-z^2}$

in the region |z|=3 .

Possible Answers:

$2 \pi i$

$\frac{\pi i}{2}$

$- \frac{\pi i}{3}$

$\pi i$

Correct answer:

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{\frac{1}{z^6}}{1 - \frac{1}{z^2}} = \frac{\frac{1}{z^6}}{z^2 - 1} = \frac{1}{z^8 - z^ 6} = -\frac{1}{z^6}* \frac{1}{1-z^2}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = -\frac{1}{z^6}* \frac{1}{1-z^2} \newline = -\frac{1}{z^6} \displaystyle \sum_{k=0}^{\infty} z^{2k} \newline = - \displaystyle \sum_{k=0}^{\infty} z^{2k -6}$

The coefficient of $\frac{1}{z}$ is .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 0$ .

Therefore, by the Residue Theorem above,

$\displaystyle \int_{|z|=3} f(z) dz = 2 \pi i \text{Res}_{z=0}[\frac{1}{z^2}f(\frac{1}{z})] = 2 \pi i (0) = 0$

Hence,

$\int_{|z|=3} f(z)dz =0$

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})]$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use the Residue Theorem to evaluate the integral of

$f(z) = 5\frac{z+1}{z(z-2)}$

in the region |z|=4 .

Possible Answers:

$\frac{\pi i}{2}$

$10 \pi i$

$- 2 \pi i$

$\pi i$

$- \frac{ \pi i }{2}$

Correct answer:

$10 \pi i$

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) =5 \frac{1}{z^2} \frac{\frac{1}{z} + 1}{ \frac{1}{z} - 2 \frac{1}{z}} = 5\frac{\frac{1}{z} +1}{1 - 2 z} = \frac{5}{z} (\frac{1}{1 - 2z}) + \frac{5}{1 - 2z}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=4 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{5}{z} (\frac{1}{1 - 2z}) + \frac{5}{1 - 2z} \newline = \frac{5}{z} \displaystyle \sum_{k=0}^{\infty} 2^{k} z^k + 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^{k-1} + 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle \sum_{k=0}^{\infty} 2^{k}( z^{k-1} + z^k)$

The coefficient of $\frac{1}{z}$ is 5 *2^0 = 5 .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 5$ .

Therefore, by the Residue Theorem above,

$\displaystyle \int_{|z|=3} f(z) dz = 2 \pi i \text{Res}_{z=0}[\frac{1}{z^2}f(\frac{1}{z})] = 2 \pi i (5) = 10 \pi i$

Hence,

$\int_{|z|=3} f(z)dz = 10 \pi i$

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Example Question #51 : Complex Analysis

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})]$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use the Residue Theorem to evaluate the integral of

$f(z) = \frac{1}{1+z^3}$

in the region |z|=3 .

Possible Answers:

$4 \pi i$

$2 \pi i$

$- \frac{\pi i }{2}$

$- \pi i$

Correct answer:

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{1}{1+\frac{1}{z^3}} = \frac{1}{z^2+z^{-1}} = \frac{ z}{ 1 + z^3 }$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{z}{1+z^2} \newline = z \displaystyle \sum_{k=0}^{\infty} (-1)^k z^{3k} \newline = \displaystyle \sum_{k=0}^{\infty} (-1)^k z^{3k+1}$

The coefficient of $\frac{1}{z}$ is .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 0$ .

Therefore, by the Residue Theorem above,

$\displaystyle \int_{|z|=2} f(z) dz = 2 \pi i \text{Res}_{z=0}[\frac{1}{z^2}f(\frac{1}{z})] = 2 \pi i (0) = 0$

Hence,

$\int_{|z|=2} f(z)dz =0$

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Example Question #2 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})]$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use the Residue Theorem to evaluate the integral of

$f(z) = \frac{2}{z}$

in the region |z|=4 .

Possible Answers:

$\pi i$

$- 2 \pi i$

$- \frac{\pi i }{4}$

$4 \pi i$

$\frac{\pi i}{2}$

Correct answer:

$4 \pi i$

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{2}{\frac{1}{z}} = \frac{1}{z^2} *2 z = \frac{2}{z}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe, the coefficient of $\frac{2}{z}$ is .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 2$ .

Therefore, by the Residue Theorem above,

$\displaystyle \int_{|z|=2} f(z) dz = 2 \pi i \text{Res}_{z=0}[\frac{1}{z^2}f(\frac{1}{z})] = 2 \pi i (2) = 4 \pi i$

Hence,

$\int_{|z|=2} f(z)dz =4 \pi i$

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Example Question #2 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})]$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use the Residue Theorem to evaluate the integral of

$f(z) = \frac{7z-3}{z(z-1)}$

in the region |z|=3 .

Possible Answers:

$4 \pi i$

$\8 \pi i$

$14 \pi i$

$2 \pi i$

$- \frac{\pi i }{4}$

Correct answer:

$14 \pi i$

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{7 \frac{1}{z} - 3}{ \frac{1}{z} (\frac{1}{z} - 1)} = \frac{7 \frac{1}{z} -3 }{ 1 -z} = \frac{7}{z (1-z)} - \frac{3}{1 -z}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{7}{z (1-z)} - \frac{3}{1 -z} \newline = \frac{7}{z}\displaystyle \sum_{k=0}^{\infty} z^{k} - 3\displaystyle \sum_{k=0}^{\infty} z^{k} \newline = 7 \displaystyle \sum_{k=0}^{\infty} z^{k-1} - 3 \displaystyle \sum_{k=0}^{\infty} z^{k}$

The coefficient of $\frac{1}{z}$ is .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 7$ .

Therefore, by the Residue Theorem above,

$\displaystyle \int_{|z|=2} f(z) dz = 2 \pi i \text{Res}_{z=0}[\frac{1}{z^2}f(\frac{1}{z})] = 2 \pi i (7) = 14 \pi i$

Hence,

$\int_{|z|=2} f(z)dz =14 \pi i$

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Example Question #1 : Residue Theory

Find the residue of the function

$f(z)= \frac{cot(z)}{z^6}$ .

Possible Answers:

$-\frac{1}{45}$

$\frac{1}{2}$

$- \frac{2}{945}$

$\frac{7}{9}$

$- \frac{2}{3}$

Correct answer:

$- \frac{2}{945}$

Explanation:

Observe

$f(z) \newline = \frac{cot(z)}{z^6} \newline = \frac{1}{z^6} (\frac{1}{z} - \frac{1}{3} z - \frac{1}{45} z^3 - \frac{2}{945} z^5 - \cdots) \newline = \frac{1}{z^7} - \frac{1}{3} \frac{1}{z^5} - \frac{1}{45} \frac{1}{z^3} - \frac{2}{945} \frac{1}{z} - \cdots$

The coefficient of $\frac{1}{z}$ is $- \frac{2}{945}$ .

Thus,

$\text{Res}_{z=0}[f(z)] = \text{Res}_{z=0}[\frac{cot(z)}{z^6}] = - \frac{2}{945}$ .

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Example Question #1 : Residue Theory

Find the residue at z=0 of

$f(z) = \cos (\frac{1}{z})$ .

Possible Answers:

$\pi$

$\frac{1}{4}$

Correct answer:

Explanation:

Let $u = \frac{1}{z}$ .

Observe,

$f(z) \newline = \cos (\frac{1}{z}) \newline =\cos (u) \newline = \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^{n}}{(2n)!} u^{2n} \newline = \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^{n}}{(2n)!} z^{-2n}$

The coefficient of $\frac{1}{z}$ is since there is no $\frac{1}{z}$ term in the sum.

Thus,

$\text{Res}_{z=0}[f(z))] = \text{Res}_{z=0}[z \cos (\frac{1}{z})] = 0$

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Complex Analysis : Residue Theory

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Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #2 : Residue Theory

Example Question #1 : Residue Theory

Example Question #51 : Complex Analysis

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